Chemical Bonding I:

Basic Concepts

Chapter 9

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2

Valence electrons are the outer shell electrons of an

atom. The valence electrons are the electrons that

participate in chemical bonding.

1A 1 ns1

2A 2 ns2

3A 3 ns2np1

4A 4 ns2np2

5A 5 ns2np3

6A 6 ns2np4

7A 7 ns2np5

Group # of valence e- e- configuration

3

Lewis Dot Symbols for the Representative Elements &

Noble Gases

4

Li + F Li+ F -

The Ionic Bond

1s22s1 1s22s22p5 1s2 1s22s22p6

[He] [Ne]

Li Li+ + e-

e- + F F -

F - Li+ + Li+ F -

LiF

Ionic bond: the electrostatic force that holds ions together in an

ionic compound.

Example 9.1

Use Lewis dot symbols to show the formation of aluminum

oxide (Al2O3).

The mineral corundum (Al2O3).

Example 9.1

Strategy We use electroneutrality as our guide in writing

formulas for ionic compounds, that is, the total positive charges

on the cations must be equal to the total negative charges

on the anions.

Solution According to Figure 9.1, the Lewis dot symbols of Al

and O are

Because aluminum tends to form the cation (Al3+) and oxygen

the anion (O2) in ionic compounds, the transfer of electrons is

from Al to O. There are three valence electrons in each Al

atom; each O atom needs two electrons to form the O2 ion,

which is isoelectronic with neon.

Example 9.1

Thus, the simplest neutralizing ratio of Al3+ to O2 is 2:3; two

Al3+ ions have a total charge of +6, and three O2 ions have a

total charge of 6. So the empirical formula of aluminum oxide is Al2O3, and the reaction is

Check Make sure that the number of valence electrons (24) is

the same on both sides of the equation. Are the subscripts in

Al2O3 reduced to the smallest possible whole numbers?

12

A covalent bond is a chemical bond in which two or more

electrons are shared by two atoms.

Why should two atoms share electrons?

F F +

7e- 7e-

F F

8e- 8e-

F F

F F

Lewis structure of F2

lone pairs lone pairs

lone pairs lone pairs

single covalent bond

single covalent bond

13

8e-

H H O + + O H H O H H or

2e- 2e-

Lewis structure of water

Double bond two atoms share two pairs of electrons

single covalent bonds

O C O or O C O

8e- 8e- 8e- double bonds

Triple bond two atoms share three pairs of electrons

N N

8e- 8e-

N N

triple bond

or

14

Lengths of Covalent Bonds

Bond Lengths

Triple bond < Double Bond < Single Bond

16

H F F H

Polar covalent bond or polar bond is a covalent

bond with greater electron density around one of the

two atoms

electron rich

region electron poor

region e- rich e- poor

d+ d-

17

Electronegativity is the ability of an atom to attract

toward itself the electrons in a chemical bond.

Electron Affinity - measurable, Cl is highest

Electronegativity - relative, F is highest

X (g) + e- X-(g)

18

The Electronegativities of Common Elements

Bond Polarity

The difference in

electronegativity

between two bonded

atoms can be used to

determine the type of

bond. As a rule of

thumb:

EN Type of Bond

Example 9.2

Classify the following bonds as ionic, polar covalent, or

covalent:

(a) the bond in HCl

(b) The bond in KF

(c) the CC bond in H3CCH3

EN values:

H = 2.1 Cl= 3.0

K = 0.8 F = 4.0

C= 2.5

24

1. Draw skeletal structure of compound showing

what atoms are bonded to each other. Put least

electronegative element in the center.

2. Count total number of valence e-. Add 1 for

each negative charge. Subtract 1 for each

positive charge.

3. Complete an octet for all atoms except

hydrogen.

4. If structure contains too many electrons, form

double and triple bonds on central atom as

needed.

Writing Lewis Structures for Molecular Compounds/

Polyatomic Ions

Example 9.3

Write the Lewis structure for nitrogen trifluoride (NF3) in which

all three F atoms are bonded to the N atom.

NF3 is a colorless, odorless, unreactive

gas.

Example 9.3

Solution We follow the preceding procedure for writing Lewis

structures.

Step 1: The N atom is less electronegative than F, so the

skeletal structure of NF3 is

Step 2: The outer-shell electron configurations of N and F

are 2s22p3 and 2s22p5, respectively. Thus, there are

5 + (3 7), or 26, valence electrons to account for in

NF3.

Example 9.3

Step 3: We draw a single covalent bond between N and each

F, and complete the octets for the F atoms. We place

the remaining two electrons on N:

Because this structure satisfies the octet rule for all the atoms,

step 4 is not required.

Check Count the valence electrons in NF3 (in bonds and in

lone pairs). The result is 26, the same as the total number of

valence electrons on three F atoms (3 7 = 21) and one N

atom (5).

Example 9.4

Write the Lewis structure for nitric acid (HNO3) in which the

three O atoms are bonded to the central N atom and the

ionizable H atom is bonded to one of the O atoms.

HNO3 is a strong electrolyte.

Example 9.4

Solution We follow the procedure already outlined for writing

Lewis structures.

Step 1: The skeletal structure of HNO3 is

Step 2: The outer-shell electron configurations of N, O, and H

are 2s22p3, 2s22p4, and 1s1, respectively. Thus, there

are 5 + (3 6) + 1, or 24, valence electrons to

account for in HNO3.

Example 9.4

Step 3: We draw a single covalent bond between N and each

of the three O atoms and between one O atom and

the H atom. Then we fill in electrons to comply with

the octet rule for the O atoms:

Step 4: We see that this structure satisfies the octet rule for

all the O atoms but not for the N atom. The N atom

has only six electrons. Therefore, we move a lone

pair from one of the end O atoms to form another

bond with N.

Example 9.4

Now the octet rule is also satisfied for the N atom:

Check Make sure that all the atoms (except H) satisfy the octet

rule. Count the valence electrons in HNO3 (in bonds and in

lone pairs). The result is 24, the same as the total number of

valence electrons on three O atoms (3 6 = 18), one N atom

(5), and one H atom (1).

Example 9.5

Write the Lewis structure for the carbonate ion ( ).

Example 9.5

Solution We follow the preceding procedure for writing Lewis

structures and note that this is an anion with two negative

charges.

Step 1: We can deduce the skeletal structure of the

carbonate ion by recognizing that C is less

electronegative than O. Therefore, it is most

likely to occupy a central position as follows:

Example 9.5

Step 2: The outer-shell electron configurations of C and O

are 2s22p2 and 2s22p4, respectively, and the ion itself

has two negative charges. Thus, the total number of

electrons is 4 + (3 6) + 2, or 24.

Step 3: We draw a single covalent bond between C and each

O and comply with the octet rule for the O atoms:

This structure shows all 24 electrons.

Example 9.5

Step 4: Although the octet rule is satisfied for the O atoms, it is

not for the C atom. Therefore, we move a lone pair

from one of the O atoms to form another bond with C.

Now the octet rule is also satisfied for the C atom:

Check Make sure that all the atoms satisfy the octet rule.

Count the valence electrons in (in chemical bonds and in

lone pairs). The result is 24, the same as the total number of

valence electrons on three O atoms (3 6 = 18), one C atom

(4), and two negative charges (2).

48

A resonance structure is one of two or more Lewis structures

for a single molecule that cannot be represented accurately by

only one Lewis structure.

O O O + -

O O O + -

Example 9.8

Draw three resonance structures for the molecule nitrous oxide,

N2O (the atomic arrangement is NNO).

Indicate formal charges.

Rank the structures in their relative importance to the overall

properties of the molecule.

Example 9.8

Strategy The skeletal structure for N2O is

We follow the procedure used for drawing Lewis structures and

calculating formal charges in Examples 9.5 and 9.6.

Solution The three resonance structures are

Example 9.8

We see that all three structures show formal charges.

Structure (b) is the most important one because the negative

charge is on the more electronegative oxygen atom.

Structure (c) is the least important one because it has a larger

separation of formal charges. Also, the positive charge is on

the more electronegative oxygen atom.

Check Make sure there is no change in the positions of the

atoms in the structures. Because N has five valence electrons

and O has six valence electrons, the total number of valence

electrons is 5 2 + 6 = 16. The sum of formal charges is zero

in each structure.

52

Exceptions to the Octet Rule

The Incomplete Octet

H H Be Be 2e-

2H 2x1e-

4e-

BeH2

BF3

B 3e- 3F 3x7e-

24e-

F B F

F

3 single bonds (3x2) = 6

9 lone pairs (9x2) = 18

Total = 24

53

Lewis Dot Symbols for the Representative Elements &

Noble Gases

54

Exceptions to the Octet Rule

Odd-Electron Molecules

N 5e- O 6e-

11e-

NO N O

The Expanded Octet (central atom with principal quantum number n > 2)

SF6

S 6e- 6F 42e-

48e- S

F

F

F

F F

F

6 single bonds (6x2) = 12

18 lone pairs (18x2) = 36

Total = 48

Example 9.9

Draw the Lewis structure for aluminum triiodide (AlI3).

AlI3 has a tendency to

dimerize or form two units as

Al2I6.

Example 9.9

Strategy We follow the procedures used in Examples 9.5 and

9.6 to draw the Lewis structure and calculate formal charges.

Solution The outer-shell electron configurations of Al and I are

3s23p1 and 5s25p5, respectively. The total number of valence

electrons is 3 + 3 7 or 24. Because Al is less electronegative

than I, it occupies a central position and forms three bonds with

the I atoms:

Note that there are no formal charges on the Al and I atoms.

Example 9.9

Check Although the octet rule is satisfied for the I atoms, there

are only six valence electrons around the Al atom.

Thus, AlI3 is an example of the incomplete octet.

Example 9.10

Draw the Lewis structure for phosphorus pentafluoride (PF5), in

which all five F atoms are bonded to the central P atom.

PF5 is a reactive

gaseous compound.

Example 9.10

Strategy Note that P is a third-period element. We follow the

procedures given in Examples 9.5 and 9.6 to draw the Lewis

structure and calculate formal charges.

Solution The outer-shell electron configurations for P and F

are 3s23p3 and 2s22p5, respectively, and so the total number of

valence electrons is 5 + (5 7), or 40.

Phosphorus, like sulfur, is a third-period element, and therefore

it can have an expanded octet.

Example 9.10

The Lewis structure of PF5 is

Note that there are no formal charges on the P and F atoms.

Check Although the octet rule is satisfied for the F atoms,

there are 10 valence electrons around the P atom, giving it an

expanded octet.

Example 9.11

Draw a Lewis structure for the sulfate ion in which all

four O atoms are bonded to the central S atom.

Example 9.11

Strategy Note that S is a third-period element. We follow the

procedures given in Examples 9.5 and 9.6 to draw the Lewis

structure and calculate formal charges.

Solution The outer-shell electron configurations of S and O

are 3s23p4 and 2s22p4, respectively.

Step 1: The skeletal structure of is

Example 9.11

Step 2: Both O and S are Group 6A elements and so have

six valence electrons each. Including the two

negative charges, we must therefore account for a

total of 6 + (4 6) + 2, or 32, valence electrons in

Step 3: We draw a single covalent bond between all the

bonding atoms:

Example 9.11

Next we show formal charges on the S and O atoms:

Note that we can eliminate some of the formal charges for

by expanding the S atoms octet as follows:

Example 9.11

The question of which of these two structures is more

important, that is, the one in which the S atom obeys the octet

rule but bears more formal charges or the one in which the S

atom expands its octet, has been the subject of some debate

among chemists. In many cases, only elaborate quantum

mechanical calculations can provide a clearer answer.

At this stage of learning, you should realize that both

representations are valid Lewis structures and you should be

able to draw both types of structures. One helpful rule is that in

trying to minimize formal charges by expanding the central

atoms octet, only add enough double bonds to make the formal charge on the central atom zero.

Example 9.11

Thus, the following structure would give formal charges on

S(2) and O(0) that are inconsistent with the electronegativities of these elements and should therefore not be included to

represent the ion.

Example 9.12

Draw a Lewis structure of the noble gas compound xenon

tetrafl uoride (XeF4) in which all F atoms are bonded to the

central Xe atom.

Example 9.12

Strategy Note that Xe is a fifth-period element. We follow the

procedures in Examples 9.5 and 9.6 for drawing the Lewis

structure and calculating formal charges.

Solution

Step 1: The skeletal structure of XeF4 is

Step 2: The outer-shell electron configurations of Xe and F

are 5s25p6 and 2s22p5, respectively, and so the total

number of valence electrons is 8 + (4 7) or 36.

Example 9.12

Step 3: We draw a single covalent bond between all the

bonding atoms. The octet rule is satisfied for the F

atoms, each of which has three lone pairs. The sum

of the lone pair electrons on the four F atoms (4 6)

and the four bonding pairs (4 2) is 32. Therefore,

the remaining four electrons are shown as two lone

pairs on the Xe atom:

We see that the Xe atom has an expanded octet.

There are no formal charges on the Xe and F atoms.

Example 9.13

Strategy

Keep in mind that bond breaking is an energy absorbing

(endothermic) process and bond making is an energy releasing

(exothermic) process.

Therefore, the overall energy change is the difference between

these two opposing processes, as described by Equation (9.3).